Use de moivres theorem to show that tan 3theta=something

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Homework Help Overview

The discussion revolves around using de Moivre's theorem to demonstrate the identity for tan 3θ. Participants are exploring the relationship between trigonometric functions and their representations through complex numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to express tan 3θ in terms of sine and cosine, while others question how de Moivre's theorem applies to the problem. There are discussions about converting terms and simplifying expressions, with some participants expressing confusion about the process of eliminating cosine terms.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and seeking clarification on specific steps. Some guidance has been offered regarding the use of identities and the application of de Moivre's theorem, but no consensus has been reached on the best approach to take.

Contextual Notes

Participants note the challenge of transforming the expression entirely into tangent terms and express uncertainty about the role of de Moivre's theorem in the context of proving the identity. There is mention of communication with the lecturer regarding the approach to take.

blueyellow

Homework Statement




use de moivres theorem to show that
tan 3theta=(3tan theta -tan^3 theta)/(1-3tan^2 theta)

Homework Equations





The Attempt at a Solution


tan 3theta=sin3theta/cos3theta
=(3sin theta - 4 sin^3theta)/(4cos^3theta -3cos theta)

so then i divided everything by cos, and som of the stuff turned into tans, but i can't get it all to turn into tans
 
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blueyellow said:

Homework Statement




use de moivres theorem to show that
tan 3theta=(3tan theta -tan^3 theta)/(1-3tan^2 theta)

Homework Equations





The Attempt at a Solution


tan 3theta=sin3theta/cos3theta
=(3sin theta - 4 sin^3theta)/(4cos^3theta -3cos theta)

so then i divided everything by cos, and som of the stuff turned into tans, but i can't get it all to turn into tans

I don't see how de Moivre's Formula enters into this problem. It seems to me to be a straightforward exercise in proving a trig identity. Although it can be done by converting the tan terms to sine/cosine, it's probably easier to leave things in terms of the tangent.

The identities that you will need are:
tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{1 - tan(\alpha)tan(\beta)}
tan(2\alpha) = \frac{2tan(\alpha)}{1 - tan^2(\alpha)}
 
If you really have to use De Moivre's theorem just consider

(cosθ + isinθ)3
 
rock.freak667 said:
If you really have to use De Moivre's theorem just consider

(cosθ + isinθ)3

but that's exactly what i considered and by equating real and imaginary coefficients, i ended up with what i wrote in my first post
and then i don't know how to proceed
any help from anyone would b much appreciated
 
wel, i emailed the lecturer and was told to just divide top and bottom by cos^3 theta
still don't see how that gets rid of the cos's tho
if u hav any idea, pls speak
thanks
 
[Im(cos theta + i sin theta)^3]/[Re(cos theta + i sin theta)^3]

=[3cos^2 theta sin theta - sin^3 theta]/[cos^3 theta - 3 cos theta sin^2 theta]

but how exactly did they go from one step to the next? and how did they get rid of the 'Im' and 'Re'?
 
Does it help that (cos(θ)+ i sin(θ))3 = (cos(3θ)+ i sin(3θ)) ?
 

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